Step 1: Understanding the Concept:
According to Snell's law in vector form, the tangential component of the product of the refractive index and the unit vector of light remains constant across an interface. Step 2: Key Formula or Approach:
Snell's Law: $\mu_1 \sin i = \mu_2 \sin r$.
This implies $\mu_1 |\hat{e} \times \hat{n}| = \mu_2 |\hat{r} \times \hat{n}|$, where $\hat{n}$ is the normal vector, $\hat{e}$ is incident unit vector, and $\hat{r}$ is refracted unit vector. Step 3: Detailed Explanation:
From the problem figure and standard convention, the normal $\hat{n}$ is along the y-axis ($\hat{j}$).
The incident vector $\vec{e} = 3\hat{i} - 2\hat{j}$.
Unit vector $\hat{e} = \frac{3\hat{i} - 2\hat{j}}{\sqrt{3^2 + (-2)^2}} = \frac{3\hat{i} - 2\hat{j}}{\sqrt{13}}$.
The refracted vector $\vec{r} = c\hat{i} - 4\hat{j}$.
Unit vector $\hat{r} = \frac{c\hat{i} - 4\hat{j}}{\sqrt{c^2 + (-4)^2}} = \frac{c\hat{i} - 4\hat{j}}{\sqrt{c^2 + 16}}$.
Snell's Law requires matching the magnitudes of the tangential components (x-axis components):
\[ \mu_1 (\hat{e}_x) = \mu_2 (\hat{r}_x) \]
Given $\mu_1 = 1$ and $\mu_2 = 1.5 = \frac{3}{2}$.
\[ 1 \times \left(\frac{3}{\sqrt{13}}\right) = \frac{3}{2} \times \left(\frac{c}{\sqrt{c^2 + 16}}\right) \]
\[ \frac{2}{\sqrt{13}} = \frac{c}{\sqrt{c^2 + 16}} \]
Squaring both sides:
\[ \frac{4}{13} = \frac{c^2}{c^2 + 16} \]
\[ 4(c^2 + 16) = 13c^2 \]
\[ 4c^2 + 64 = 13c^2 \]
\[ 9c^2 = 64 \implies c^2 = \frac{64}{9} \]
\[ c = \frac{8}{3} \approx 2.67 \] Step 4: Final Answer:
The value of c is 2.6.
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